If X is a geometric random variable, show analytically that P(X = n+k|X >n) = P(X = k) .Give a verbal argument using the interpretation of a geometricrandom variable as to why the preceding equation is true.A geometric random variable is a discrete random variable with aprobability mass function defined as follows:p(n)=(1-p)n-1p, n=1,2,....where 0 < p < 1.

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usamakld99 usamakld99 · Answer 1 · 2019-11-11T17:35:40+01:00

Answer:

P[(X=n+k)] ∩ X>n)] =P[X=K]

Step-by-step explanation:

If X is a geometric random variable then

for success probability = p

so for failure q= 1-p

Now as given

[tex]P_{x}(n)=(1-p)^{x-1}P_{}[/tex]

Now for the P parameter we have

x∈{1,2,3,....∞}

[tex]P [X=K] = (1-P)^{K-1}P_{}[/tex]

[tex]P [X=n+K] = (1-P)^{n+K-1}P_{}[/tex]

[tex]p[\frac{X=n+K}{X>h}]=\frac{[(X=n+K)n,X>n]}{P(X>n)}[/tex]

[tex]P(X>n)=\sum_{i=n+1}^{\infty}(1-P)^{i-1}P^{} \\[/tex]

[tex]P(X>n)=P\sum_{i=n+1}^{\infty}(1-P)^{i-1}^{}[/tex]

[tex]P(X>n)=P\frac{1-P}{1-(1-P)^{n} } =(1-P)^{n}[/tex]

P[(X=n+k)] ∩ X>n)] = P(X=n+K)

P[(X=n+k)] ∩ X>n)] = [tex]\frac{(1-P)^{n+k-1}P }{(1-P)^{n} }[/tex]

P[(X=n+k)] ∩ X>n)] = [tex](1-P)^{k-1} .P[/tex]

P[(X=n+k)] ∩ X>n)] =P[X=K]